# How do I rewrite sin(x^1/2)?

1. May 8, 2010

### twotaileddemon

1. The problem statement, all variables and given/known data

I would like to know how to rewrite sin[x^(1/2)]? I have to do a laplace transform of a certain equation but I need to get it into a more usable form...

2. Relevant equations

3. The attempt at a solution

I was thinking of using the binomial theorem but 1/2 is neither a even or odd number so I can't use it?

2. May 8, 2010

### rock.freak667

I don't think there is a way to get in a more usable form. Was your equation to solve directly involving sin(√x)?

3. May 8, 2010

### twotaileddemon

essentially, yes, where x = at. and

f(t) = sin(√at)

I didn't know how to solve it because there was a t^(1/2) term X_X. Do you have any ideas?

Can I just use the formula L(f) = F(s) = $$\int$$f(t)e-st dt where the integration runs from 0 to infinity, even though I have a t^(1/2) term?

Last edited: May 8, 2010
4. May 8, 2010

### gabbagabbahey

Of course. That is the definition of the Laplace transform after all (you are missing your integration limits though)

5. May 9, 2010

### twotaileddemon

can you have a fourier transform with a complex number in it? The answer I got has a complex number in it...

6. May 9, 2010

### gabbagabbahey

If $a$ is real, your answer should be too. When you take the Laplace transform of a real-valued function, you get a real-valued function.

Maybe you should show us your calculation...

7. May 9, 2010

### Hellabyte

You could try 'u' substitution

8. May 9, 2010

### twotaileddemon

well I'm not sure how to do all the symbols and stuff very well, but basically, I did

$$\int$$sin[(at)1/2]e-stdt

substituted 1/(2i) * ( exp[i(at)1/2] - exp[-i(at)1/2] ) for the sine term, integrated with limits between 0 and infinity, and eventually got

-i / s

as my answer, where i is an imaginary number.

9. May 9, 2010

### gabbagabbahey

See my sig

10. May 9, 2010

### twotaileddemon

Okay!

1/(2i) * [ $$\int$$ exp[i(at)1/2] - st] dt - $$\int$$ exp[-i(at)1/2] - st] dt ] =

1/(2i) * [ 1 / [ i(a / 2(at)1/2) - s] - 1 / [ -i(a / 2(at)1/2) - s] ]

*note at this point I plugged in 0 for t for the lower limit and infinity for the upper limit of t from integration*
= 1/(2i) * [1/s + 1/s]
= 1/(2i) * [2/s]
= 1/is = -i/s

11. May 9, 2010

### gabbagabbahey

You seem to be claiming that

$$\int e^{h(t)}dt=\frac{e^{h(t)}}{h'(t)}$$

Is that really true? (Use the product rule to differentiate $$\frac{e^{h(t)}}{h'(t)}$$ and see if you get $e^{h(t)}[/tex]) Instead, start with the substitution [itex]u=\sqrt{at}$ and then complete the square on the exponent...

12. May 9, 2010

### twotaileddemon

I did the substitution and now I have an exponent equal to

exp(iu - su2/a)

and another equal to

exp(-iu - su2/a)

can I complete the square with this and solve it?

13. May 9, 2010

### gabbagabbahey

Yes, start by completing the square...

14. May 9, 2010

### twotaileddemon

ok well..

exp(iu - su2/a)

= exp[-(su2/a - iu)]
= exp[-(s/a)(u2 - i(a/s)u)]
= exp[-(s/a)(u-(ia/2s)2 - (i2/(4(s/a))]
= exp[-(s/a)(u-(ia/2s)2 + a /(4s)]
...?

15. May 9, 2010

### vela

Staff Emeritus
Up to here it looks okay
Your parentheses don't match up here, but what you actually have is probably right.

Note that ea/(4s) is a constant, so you can pull it out of the integral.

16. May 9, 2010

### gabbagabbahey

Right, now $$e^{-\frac{s}{a}\left(u-\frac{ia}{2s}\right)^2+\frac{a}{4s}}=e^{\frac{a}{4s}}e^{-\frac{s}{a}\left(u-\frac{ia}{2s}\right)^2}$$ and $$e^{\frac{a}{4s}}$$ is a constant and can be pulled out of the integral.